Question

Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Hint:

The configuration always creates a right triangle with the larger radius as the hypotenuse (\(R\)), the smaller radius as one side (\(r\)), and half the chord length as the other side (\(L/2\)).
Formula: \(L = 2\sqrt{R^2 - r^2}\).

Answer 1

Given:
Two concentric circles have radii:
Larger circle radius \(R = 5\) cm
Smaller circle radius \(r = 4\) cm

We need to find:
The length of the chord of the larger circle that touches the smaller circle.

Step 1: Understand the geometry
Since the circles are concentric, the distance between their centres is 0.
A chord of the larger circle touches the smaller circle, so it is tangent to the smaller circle.

The perpendicular distance from the centre to this chord equals the radius of the smaller circle: \[ d = r = 4\ \text{cm} \]

Step 2: Use chord length formula
For a circle of radius \(R\), the length of a chord at distance \(d\) from the center is: \[ \text{Chord length} = 2\sqrt{R^2 - d^2} \]

Step 3: Substitute the values
\[ = 2\sqrt{5^2 - 4^2} \] \[ = 2\sqrt{25 - 16} \] \[ = 2\sqrt{9} \] \[ = 2 \times 3 = 6\ \text{cm} \]

Final Answer:
The length of the chord is: \[ \boxed{6\ \text{cm}} \]